HDU 3339 In Action 最短路+01背包~

题目大意就是 你有足够多的坦克，让你用尽量多的坦克去占领电厂。有N+1各节点，0~n ，0为出发点，1~n为电厂。要想控制摧毁电厂，战胜对方，只需要占领一般的电量即可。但是特克会耗油，所一给你m条路，让你去计算最小耗电量~而且如果坦克不能通过m条路到达任意一个电厂的话就输出'impossible'

连接：http://acm.hdu.edu.cn/showproblem.php?pid=3339

代码：dij

#include <stdio.h>
#include <string.h>
#define maxn 0x5fffffff
int map[105][105];
int dis[105];
void inint(int n)
{
    int j,i;
    for(i = 0;i <= n;i++)
    {
        for(j = 0;j <= n;j++)
        if(i != j)map[i][j] = maxn;
        else map[i][j] = 0;
    }
}
void disj(int n)
{
    int vis[105] = {0};
    int pre,i,j,k,min;
    pre = 0;
    for(i = 0;i <= n;i++)
    dis[i] = map[pre][i];
    vis[pre] = 1;

    for(k = 1;k <= n;k++)
    {
        for(i = 0;i <= n;i++)
        {
            if(!vis[i] && dis[i] > dis[pre]+map[pre][i])
            dis[i] = dis[pre]+map[pre][i];
        }
        min = maxn;
        for(i = 0;i <= n;i++)
        if(min > dis[i] && !vis[i])
        min = dis[i],pre = i;

        vis[pre] = 1;
    }
}
int main()
{
    int T,n,m,i,j,a,b,c,sum,pow[105],spow,f[20005];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        inint(n);
        for(i = 0;i < m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(c < map[a][b])
            map[a][b] = map[b][a] = c;
        }

        disj(n);
        spow = sum = 0;
        for(i = 1;i <= n;i++)
        scanf("%d",pow+i),spow+=pow[i];
        int leap = 0;
        for(i = 1;i <= n;i++)
        {
            if(dis[i] != maxn)
            sum += dis[i];
            else
            leap = 1;
        }
        if(leap)
        puts("impossible");
        else
        {
            memset(f,0,sizeof(f));
            for(i = 1;i <= n;i++)
            {
                for(j = sum;j >= dis[i];j--)
                {
                    if(f[j] < f[j-dis[i]]+pow[i])
                    f[j] = f[j-dis[i]]+pow[i];
                }
            }
            for(i = 0;i <= sum;i++)
            if(f[i] > spow/2)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

floyd

#include <stdio.h>
#include <string.h>
#define maxn 10000050
int map[105][105];
int dis[105];
void inint(int n)
{
    int j,i;
    for(i = 0;i <= n;i++)
    {
        for(j = 0;j <= n;j++)
        if(i != j)map[i][j] = maxn;
        else map[i][j] = 0;
    }
}

int main()
{
    int T,n,m,i,j,k,a,b,c,sum,pow[105],spow,f[100005];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        inint(n);
        for(i = 0;i < m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(c < map[a][b])
            map[a][b] = map[b][a] = c;
        }

        spow = sum = 0;
        for(i = 1;i <= n;i++)
        scanf("%d",pow+i),spow+=pow[i];

        for(k = 0;k <= n;k++)
        {
            for(i = 0;i <= n;i++)
            {
                for(j = 0;j <= n;j++)
                if(map[i][j] > map[i][k]+map[k][j])
                map[i][j] = map[i][k]+map[k][j];
            }
        }

        for(i = 1;i <= n;i++)
        {
            dis[i] = map[0][i];
            if(dis[i] != maxn)
            sum += dis[i];
        }

        memset(f,0,sizeof(f));
        for(i = 1;i <= n;i++)
        {
            for(j = sum;j >= dis[i];j--)
            if(f[j] < f[j-dis[i]] + pow[i])
            f[j] = f[j-dis[i]] + pow[i];
        }

        for(i = 0;i <= sum;i++)
        {
            if(f[i] >= spow/2+1)
            {
                printf("%d\n",i);
                break;
            }
        }
        if(i > sum)
        puts("impossible");
    }
    return 0;
}